Question

Use the given information to write the standard equation of the circle.

a circle that has (-2,4) and (4,2) as endpoints of a diameter.

Answer 1

so you said in your messages that you got 40? (remember that there's a square root sign so it's sqrt(40))

that's correct for the diameter. the radius is one half of that, so (1/2) * sqrt(40). feel free to keep going w/ the midpoint.

Answer 2

uh i dont think i got the right calculator for this cs i just get decimals

Answer 3

for these problems, leave the distance formula result in terms of radicals only. rounding to a decimal will result in lost accuracy.

in this case, you got sqrt(40) so just leave it as sqrt(40).

the diameter is sqrt(40) so the radius is (1/2)sqrt(40).

Answer 4

notice how the standard form of a circle has r^2 so the radical will get squared and the radical part will disappear leaving you with a whole number

Answer 5

this is confusing

Answer 6

this is just an example

let's suppose you get sqrt(2) as the answer for your distance formula calculation. just leave it as sqrt(2) instead of rounding it. square roots, if they're not perfect squares, have a decimal part that just goes on forever, so you lose accuracy if you round.

if you get d = sqrt(2), leave it like that instead of rounding it to 1.41.

if your diameter is sqrt(2), then the radius is (1/2)sqrt(2)

now, going back to the standard form equation y = (x-h)^2 + (y-k)^2 = r^2, if we plug in r = (1/2)sqrt(2) we get r^2 = (1/2)^2 * (sqrt(2))^2 = (1/4) * 2 = 1/2

notice how the square root sign disappeared, because we squared sqrt(2).

Answer 7

now going back to the problem, you got distance = sqrt(40) so just leave it like this. radius is one half the diameter so radius = (1/2)sqrt(40)

leave it like that, and move on to the midpoint calculation.

Answer 8

ok

Answer 9

i got (1,3)

Answer 10

@vocaloid is this correct?

Answer 11

yup good, so plugging everything in

\[(x-1)^{2}+(y-3)^{2}=((\frac{ 1 }{ 2 })(\sqrt{40}))^{2}\]

\[(x-1)^{2}+(y-3)^{2}=10\]

Answer 12

thnk you